cs5371 theory of computationwkhon/toc07-lectures/lecture22.pdfd-hampath is np-complete (14) in the...

CS5371Theory of Computation

Lecture 22: Complexity VII(More NP-complete Problems)

•We shall continue to look at more NP-complete problems:

–Directed Hamiltonian Path–Hamiltonian Path–SUBSET SUM–PARTITION

•Some more if we have time today

Objectives

Directed HAMPATH

Theorem: D-HAMPATH is NP-complete.

Let G be a directed graph. A directedHamiltonian path in G is a path that visitsall the vertices of G once and only once.

Let D-HAMPATH be the language

{ G,s,t| G has a directed Hamiltonianpath from s to t }

D-HAMPATH is NP-complete (2)

Proof: First, D-HAMPATH is in NP (easyto show). Then, we show it is NP-completeby reduction from 3SAT.

To determine if Fis in 3SAT, we shallconstruct G with two special vertices s, tsuch that

F3SAT G,s,tD-HAMPATH


Let x1, x2, …, xj be the variables in the3cnf-formula F. For each variable xi, wecreate a ‘diamond-shaped structure’thatcontains a horizontal row of nodes:

xi …


Let C1, C2, …, Ck be the clauses in F. Foreach clause Cm, we create a node:

Cm

Next slide shows the global structure of G.

It shows all the elements of G and theirrelationship, except the relationship of thevariables to the clauses that contains them

C2

…

Ck

C3

C1

s

x1

…x2

…xj

tHigh-level structure of G

Diamond Structure

Each diamond structure contains ahorizontal row of nodes connected byedges running in both directions.

There are 3k+1 nodes in each row (inaddition to the two nodes on the end thatbelongs to the diamond)

Diamond StructureThe nodes in the horizontal row aregrouped into adjacent pairs, one for eachclause, with extra separator nodes next tothe pair:

…

pairsfor C1

pairsfor C2

xi

Connection with Clauses

If variable xi appears in clause Cm, we addthe following two edges from the mth pairin the xi’s diamond structure to node Cm

…

pairsfor Cm

xi…

Cm

Connection with Clauses (2)

If :xi appears in clause Cm, we add thefollowing two edges from the mth pair inthe xi’s diamond structure to node Cm

…

pairsfor Cm

xi…

Cm

Note thedirection change


After we add all the edges correspondingto each occurrence of xi or : xi in eachclause, the construction of G is finished

We claim that this is our desired reduction:

(=>) Suppose that F is satisfiable. Todemonstrate an Hamiltonian path from s tot in G, we first ignore the clause nodes


The path begins at s, goes through eachdiamond in turn, and ends up at t

To hit the horizontal nodes in a diamond,the path is either one of the following way:

… …

Left-to-Right Right-to-Left


If xi is assigned TRUE in the satisfyingassignment of F, we use Left-to-Rightmethod to traverse the correspondingdiamond. Otherwise, if xi is assignedFALSE, we use the Right-to-Left method

So far, the path covers all the nodes in Gexcept the clause nodes. We can easilyinclude them by adding detours at thehorizontal node


In each clause, we select a literal that isassigned TRUE in the satisfying assignmentof F

Suppose we select xi in clause Cm. Then, inour current path, the horizontal nodes in xiare from Left-to-Right. Also, by ourconstruction of edges in P. 11, we can seethat we can easily detour at the mth pair ofhorizontal nodes, and cover Cm


Similarly, if we select : xi in clause Cm, thenin our current path, the horizontal nodes inxi are from Right-to-Left. Also, by ourconstruction of edges in P. 12, we can seethat we can easily detour at the mth pair ofhorizontal nodes, and cover Cm

Thus, if F is satisfiable, G has a Hamiltonianpath from s to t (proof of the => direction done)


(<=) For the other direction, if G has aHamiltonian path from s to t, we shalldemonstrate a satisfying assignment for F

Firstly, we take a look at the Hamiltonianpath. If it is “normal”--- that is, visitingthe diamonds in the order from top one tothe bottom one (excluding the detours toclause nodes) --- we can obtain a satisfyingassignment as follows: (next slide)


•If the path is Left-to-Right in thediamond for xi, we assign TRUE to xi

•If the path is Right-to-Left in thediamond for xi, we assign FALSE to xi

Because each clause node is visited once inthe Hamiltonian path, by determining howthe detour is taken, we know that at leastone literal in each clause is TRUE


It remains to show is:

Hamiltonian path must be normal

We prove this by contradiction.

Suppose on the contrary that the path isnot normal. Then, the Hamiltonian pathmust have entered a clause from onediamond, but left the clause to anotherdiamond, as shown in next slide:

When Hamiltonian Path not Normal

…

…

Cm

a1 a2 a3

b


The Hamiltonian path goes from a1 to c butinstead of returning to a2, it goes to b ina different diamond

However, in our graph G, a2 is connected toat most three nodes: a1 , a3, and c (why?)

If a2 is a separator: connects to a1 and a3 only

Else: connects to a1, a3, and c only


In the current path, a1 and c have both beenvisited now. So a2 cannot find two distinctnodes (one incoming neighbor, one outgoing neighbor)that connects it to the current path

Thus, current path is not Hamiltonian !!

and a contradiction occurs (where?)

So, all Hamiltonian path from s to t must benormal. This implies that if such a pathexists, F is satisfiable (proof of <= direction done)


In conclusion, we have

F3SAT G,s,tD-HAMPATH

As it is easy to see that the above reductionfrom 3SAT to D-HAMPATH takes onlypolynomial time, D-HAMPATH is thereforeNP-complete

Undirected HAMPATH

Theorem: HAMPATH is NP-complete.

Let HAMPATH be the language

{ G,s,t| G has a (undirected) Hamiltonianpath from s to t }

HAMPATH is NP-completeWe just give a sketch of the proof:

1. First, HAMPATH is in NP (easy to check).

2. Next, to see why it is NP-complete, wereduce 3SAT to HAMPATH, using similarconstruction as we use in D-HAMPATH.

However, the graph is now undirected.Instead of using directed edges in thereduction in D-HAMPATH before, we …

HAMPATH is NP-complete (2)…replace every node u in previous graph by3 nodes, uin, umid, uout, in the new graph

A directed edge from u to v in the previousgraph is now replaced by an undirectededge joining uout and vin

This completes the reduction, and we canshow that this reduction works (exerciseat home!) and takes polynomial time

Thus, HAMPATH is NP-complete

How about HAM-CIRCUIT ?

Theorem: HAM-CIRCUIT is NP-complete.

Let HAM-CIRCUIT be the language

{ G| G has a Hamiltonian circuit }

HAMCIRCUIT is NP-complete (2)

Proof: First, HAMCIRCUIT is in NP (easyto show). Then, we show it is NP-completeby reduction from HAMPATH.

To determine if G,s,tis in HAMPATH, weconstruct G’by adding to G a new vertex v,and two edges {v,s} and {v,t}. Then it iseasy to see that:

G,s,tHAMPATH G’HAMCIRCUIT

SUBSET-SUM is NP-Complete

Theorem: SUBSET-SUM is NP-complete.

Let S be a set of positive integers.

Let SUBSET-SUM be the language

{ S,t| S has a subset whose sum is t }

SUBSET-SUM is NP-complete (2)

Proof: First, SUBSET-SUM is in NP (easyto show). We show it is NP-complete byreduction from 3SAT

Let F be a Boolean formula in 3cnf-form.Let x1, x2, …, xj be its variables and let C1,C2, …, Ck be its clauses.

We transform F into a set S of 2j+2k (verylarge) numbers, with each having j+k digits(the first j digits corresponds to variables, and thefollowing k digits corresponds to clauses)


For each variable xi, we create two numbersyi and zi, such that their ith digit (correspondingto xi) is set to 1. Also,

•if xi appears in Cm, set the (j+m)th digitof yi (which corresponds to Cr) to 1

•if :xi appears in Cm, set the (j+m)th digitof zi (which corresponds to Cm) to 1

The remaining digits of yi and zi are set to 0

Constructing the numbers in S

001000zj

101000yj

010010z2

100010y2

000001z1

010001y1

CkC2C1j321

Assume C1 = ( x1 _ : x2 _ x3) and C2 = ( x2 _ : x3 _ xj)


In addition, S contains one pair of numbers,gm and hm for each clause Cm

These two numbers are equal, with the(j+m)th digit (which corresponds to Cm) set to 1[Later, they are used as ‘fillers’to get a subset sum]

Let t = 1111 3333 [j 1s followed by k 3s]be the target number. We shall show:

F is satisfiable a subset of S adds to t


(=>) Suppose F is satisfiable. We select yi ifxi is set to TRUE. Else, we select zi

If we add the numbers selected so far,

1. Leftmost j digits are correct (why?)

2. Each of the remaining k digits isbetween 1 and 3 (why?)

Now, we can select suitable gm or hm (or both)to fill up the differences, thus getting t


(<=) On the other hand, suppose a subset ofS adds up to t. Then, exactly one of theyi or zi is in this subset (why?)

By setting xi to TRUE when yi is in thesubset, and FALSE when zi is in thesubset

We claim F is satisfied: Consider thenumbers in the subset whose (j+m)th bit(corresponding to Cm) is 1


There must be 3 such numbers (why?), soone of them must be some yi or zi (why?)

•If it is yi, it means (i) Cm contains xi,and (ii) we have assigned xi to TRUE

•If it is zi, it means (i) Cm contains : xi,and (ii) we have assigned xi to FALSE

In both cases, Cm is satisfied

Thus, all clauses are satisfied, and so is F


Now, we have shown that

F3SAT S,tSUBSET-SUM

Also, it is easy to check that the abovereduction takes polynomial time (in termsof the length of F).

Thus, SUBSET-SUM is NP-complete.

PARTITION is NP-Complete

Theorem: PARTITION is NP-complete.

Let S be a set of positive integers.

Let PARTITION be the language

{ S| S can be partitioned into two groupssuch that the sum in each group isthe same }

PARTITION is NP-complete (2)

Proof: First, PARTITION is in NP (easy toshow). Then, we show it is NP-complete byreduction from SUBSET-SUM.

To determine if S,k is in SUBSET-SUM, letX = sum of values in S. We construct S’byadding the two numbers 2X-k and X+k to S.Then it is easy to see that (why??):

S,kSUBSET-SUM S’PARTITION

Brain Teaser 1: HITTING SET

Theorem: HITTING-SET is NP-complete.

Let C be a collection of subsets of S. A setof S’is called a hitting set for C if everysubset of C has at least one element in S’.

Let HITTING-SET be the language

{ C,k| C is a collection of subsets with ahitting set of size k }

Brain Teaser 2:SUBGRAPH ISOMORPHISM

Theorem: SUBGRAPH-ISO is NP-complete

We say two graph H=(V,E) and H’=(V’,E’)are isomorphic if there exists a one-to-onefunction f: V’ V such that

{ u,v } in E if and only if { f(u),f(v) } in E’

Let SUBGRAPH-ISO be the language

{ G,H| G has a subgraph isomorphic to H}

SUBGRAPH-ISO is NP-complete

Proof: First, SUBGRAPH-ISO is in NP(easy to show). Then, we show it is NP-complete by reduction from CLIQUE.

Given G,k, we construct G’,H’as follows:

Set G’= G. Set H’= k-clique. Then it iseasy to see that:

G,kCLIQUE G’,H’SUBGRAPH-ISO

Brain Teaser 3:BOUNDED-DEG SPANTREE

Theorem: Bounded-Deg-ST is NP-complete

A spanning tree of a graph G=(V,E) is atree containing every vertex in G, andwhose edges are from E. A degree-kspanning tree is a spanning tree such thatdegree of each internal node is at most k

Let Bounded-Deg-ST be the language

{ G,k| G has a degree-k spanning tree }

Bounded-Deg-ST is NP-complete

Proof: First, Bounded-Deg-ST is in NP(easy to show). Then, we show it is NP-complete by reduction from HAMPATH

Hint: What is a degree-2 spanning tree?

A degree-2 spanning tree is a Hamiltonianpath in the graph!!!

Bounded-Deg-ST is NP-complete (2)

Now, given a graph G, we can transform Ginto G’by adding two nodes, u and v, andtwo edges, {u,s} and {v,t}. Then, we can see

G,s,tHAMPATH

G’,2Bounded-Deg-ST

So, Bounded-Deg-ST is NP-complete

Brain Teaser 4: KNAPSACK

Theorem: KNAPSACK is NP-complete.

Let S be a set of items, each item x in Shas a positive integral value v(x) and apositive integral weight w(x).

Let KNAPSACK be the language

{ S,b,k| a subset of items in S of totalweight at most b, but whose totalvalue is at least k }

KNAPSACK is NP-completeProof: First, KNAPSACK is in NP (easy toshow). Then, we show it is NP-complete byreduction from PARTITION

Let S = {s1,s2,…,sj} be a set of +ve integers.We want to construct S’, b, and k such that

SPARTITION S’,b,kKNAPSACK

The construction of S’is as follows:

KNAPSACK is NP-complete (2)For each si in S, we create xi in S’such that

w(xi) = v(xi) = si

Also, set b = k = Y/2, where Y = s1+s2+…+sj

Now, if S has a partition, then a subset ofnumbers in S adds up to Y/2. By choosingthe items of S’that corresponds to thissubset, the total weight b and the totalvalue k (why?)

Thus, SPARTITION S’,b,kKNAPSACK

KNAPSACK is NP-complete (3)

On the other hand, if a subset of items ofS’have total weight b and total value k,then the sum of the corresponding si in Swill be at most b and at least k (why?).

Since b = k = Y/2, we have the sum ofthose items in S = Y/2Thus, S’,b,kKNAPSACK SPARTITION

As the reduction can be done in polynomialtime, KNAPSACK is NP-complete.

What we have learnt

• The class P, and the class NP• Some problems in NP are the most

difficult ones in the set. We call themNP-complete problems

• SAT is NP-complete• Other problems in NP can be shown to be

NP-complete using polynomial timereduction (from what to what?)

Next Time

• Chapter 8 (not in exam) or Revision?

Decision vs Optimization• The problems we have seen so far are

called decision problems, because we wantto decide if a string is in a set or not

• For instance, in the vertex cover problem,we have a language VERTEX-COVER

{ G,k| G is a graph with a vertex coverof size k }

and our task is to decide if a string is inthe set or not

Decision vs Optimization• For most decision problems we studied,

we can define its optimization version

• E.g, in vertex cover problem, we can ask:

Given a graph G, what is the minimumsubset of vertices that can cover G?

• For the above problem, we don’t want ayes/no answer, but we want a specificcover whose size is minimum

Decision vs Optimization

• Recall: VERTEX-COVER is NP-complete Unlikely to solve its optimization (min)

version in polynomial time

• Other than give up, we may find:• Better exponential-time algorithm

• improve from O(2n) to O(1.28n)• Approximation algorithm• Randomized algorithm• Simplified version

• restrict G to be bipartite

Approximation Algorithm?• Algorithm with performance guarantee

• E.g., an algorithm that can always find acover whose size is at most k times theoptimal one

We call this a k-approximationalgorithm for the vertex cover problem

Approximation Algorithm• In general, a k-approximation algorithm

• for minimization problem guarantees asolution at most k times the optimalone; while

• for maximization problem (such asfinding a maximum clique in G)guarantees a solution at least 1/k timesthe optimal one

• smaller k = better guarantee

Approximation Algorithm• For vertex cover, can we find a good

approximation algorithm?

• Good means:• It runs fast (in polynomial time)• It has small k

• Let us try to find a 2-approximationalgorithm for vertex cover

cs5371 theory of computationwkhon/toc07-lectures/lecture22.pdfd-hampath is np-complete (14) in the...

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